Ring structure on $Ext$ and $Tor$

Question

Wikipedia says that in certain situations, $Ext^\ast_A(R,R)$ becomes a ring, such as when $A$ is an augmented $R$-algebra, but the outline is too sketchy for me to understand.

I can't find this in Weibel. Isn't $Ext^\ast_A(M,M)$ a ring (even an $\mathbb{N}$-graded algebra over center $Z(A)$) for any algebra $A$, via the splicing (concatenation) of extensions (exact sequences) $$0\to M\to E_k\to\ldots\to E_1\to M\to0\text{ and }0\to M\to E'_l\to\ldots\to E'_1\to M\to0,$$ $$\text{producing }0\to M\to E_k\to\ldots\to E_1\to E'_l\to\ldots\to E'_1\to M\to0\in Ext^{k+l}_R(M,M)?$$

A more detailed sketch or a reference would be welcome. How can this multiplication be defined when we interpret $Ext^n_A(M,M)$ as $H_n(Hom_A(P_\ast,M))$ for some projective resolution $P_\ast\to M$?

If I understand correctly, this makes the group and algebra and Lie algebra cohomology $\bigoplus_{n\geq0}Ext^n_{R[G]}(R,R)$ and $\bigoplus_{n\geq0}Ext^n_{A\otimes A^{op}}(R,R)$ and $\bigoplus_{n\geq0}Ext^n_{U(\mathfrak{g})}(R,R)$ into rings, correct?

Can some similar construction be made for $Tor$?

Answer 1

You should think of $\text{Ext}^{\bullet}_A(R, R)$ as a shadow of a more fundamental operation, namely taking derived endomorphisms $\text{RHom}_A(R, R)$ of $R$, regarded as an object in the derived category of $A$-modules. Here the ring structure is obvious; the multiplication comes from composition. More generally, there is a composition map $\text{RHom}(A, B) \times \text{RHom}(B, C) \to \text{RHom}(A, C)$, which you can write down by replacing $A, B, C$ with projective resolutions and then composing chain maps.

The situation for $\text{Tor}$ is different; you should also think of $\text{Tor}$ as a shadow of a more fundamental operation, namely taking derived tensor products $M \otimes_A^L N$, but even in the underived setting the tensor product of two modules has no reason to acquire a ring structure. One should only expect a ring structure if $M$ and $N$ are themselves already $A$-algebras, which happens e.g. if $A$ has an augmentation $A \to R$ and $M = N = R$.

Answer 2

Let $k$ be a commutative ring and let $A$ be an augmented $k$-algebra, so that in particular $k$ is an $A$-module. Fix a projective resolution $(P,d_P)$ of $k$. Then $EP=\text{End}_A(P)$ is a complex with differential given by $df = d_P f -(-1)^{|f|} f d_P$, and it has a product given by composition of maps: for endomorphisms $f$ and $g$, set $f\smile g = fg$ (the composition). If $d$ is the differential of $EP$ you can check that $$d(f\smile g) = df\smile g +(-1)^{|f|} f\smile dg $$ so $EP$ is a differential graded algebra. This means its homology, which is $\text{Ext}_A(k,k)$, is also a differential graded algebra with the product induced by $\smile$.

You can also consider a free resoluion $P= V\otimes A$ so that $\text{Hom}_A(V\otimes A,k)= \text{Hom}_k(V,k)=V^*$, and you can construct a section $i$ of the map $p : EP=\text{Hom}_A(P,P) \to \text{Hom}_A(P,k)$ given by postcomposition with the augmentation of $P\to k$ so that $ip$ is homotopic to the identity. Then you can define a product on $V^*$ by the rule $f\smile' g =p( if\smile ig)$ which induces the same product on $H(V^*) = \text{Ext}_A(k,k)$.

In both cases one can in fact produce on $\text{Ext}$ a minimal $A_\infty$-algebra structure, with $\smile$ being $m_2$.